Bash arithmetic evaluation and errexit trap

Fri 08 August 2014

In the "traps for new players" category:

count=0
things="0 1 0 0 1"

for i in $things;
do
   if [ $i == "1" ]; then
       (( count++ ))
   fi
done

echo "Count is ${count}"

Looks fine? I've probably written this many times. There's a small gotcha:

((expression))

The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION. If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to let "expression".

When you run this script with -e or enable errexit -- probably because the script has become too big to be reliable without it -- count++ is going to return 0 (post-increment) and per above stop the script. A definite trap to watch out for!